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## Homework Statement

You want to know the height of a cliff so you throw a rock off the edge with initial speed v0 = 42 m/s upwards, inclined at an angle θ = 27° with respect to the horizontal. You have a friend that records the time that it takes for the rock to hit the bottom of the canyon below. It takes 7.6 s for the rock to hit the bottom of the canyon from when you throw it.

## Homework Equations

v0x = v0*cos(theta)

v0y = v0*sin(theta)

ay = -g

ax = 0

vy(t) = v0*sin(theta) - g*t

vx(t) = v0*cos(theta)

x(t) = h + v0*sin(theta)*t - 1/2*g*t^2

y(t) = v0*cos(theta)*t

h = height of the cliff

v = velocity

v0 = initial velocity

a = acceleration

## The Attempt at a Solution

I am just confused about how to find the height of the cliff. I can't use the initial time because the initial time is zero. I tried solving for the x-distance at the final time. but I don't know how that would help. Any suggestions please?!?!